As a technician in a large pharmaceutical research firm, you need to produce 350

Question

As a technician in a large pharmaceutical research firm, you need to produce 350.mL of a potassium dihydrogen phosphate buffer solution of pH = 7.08. The pKa of H2PO4? is 7.21.

You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.

How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)

Express your answer to three significant digits with the appropriate units.

Please explain and don't skip any steps.

Explanation / Answer

Let a be the volume of KH2PO4 and b be the volume of K2HPO4 needed in liters

Moles of KH2PO4 = volume x concentration = a x 1.00 = a mol

Moles of K2HPO4 = volume x concentration = b x 1.00 = b mol

Henderson-Hasselbalch equation:

pH = pKa + log([K2HPO4/[KH2PO4]])

= pKa + log(moles of K2HPO4/moles of KH2PO4) since final volume (= 350 mL) is the same for both

7.08 = 7.21 + log(b/a)

log(b/a) = 0.74

b/a = 0.74

thus b = 0.74a

Total moles of phophate = final volume x total concentration of phosphate

= 350/1000 x 1.00 = 0.350 mol

Thus a + b = 0.350

a + 0.74a = 0.450

a = 0.258

Volume of KH2PO4 needed = a = 0.258L

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